Currently studying for qualifying exams and came across the following problem:
Give an explicit example of a ring $R$ (commutative with identity) and a surjection $\psi: M \rightarrow N$ of finitely generated $R$-modules such that $\ker(\psi)$ is not a finitely generated $R$-module.
The given hint is to let $R = \mathbb{Z}[x_1, x_2, ... ]$, however I am confused how such a construction would yield finitely generated submodules? It seems clear that we want to invoke the general strategy of picking an ideal $I$ of $R$ and defining the natural surjection $$\phi : R \rightarrow R/I$$ If we let $I = \langle x_1, x_2, ...., \rangle$ then $\ker(\phi) \cong I$ and $I$ is not finitely generated, however $R$ is also not finitely generated. Thus, we do not satisfy the problem statement? Any help on this with the given hint would be greatly appreciated!
A (commutative with identity) ring $R$ is Noetherian if and only if every ideal of $R$ is finitely generated (as a module over $R$ or as an ideal, it's the same).
A ring $R$ is Noetherian if and only if every submodule of every finitely generated $R$-module is finitely generated.
Thus your example requires a ring that's not Noetherian and the simplest example is $R=\mathbb{Z}[x_1,x_2,\dotsc]$, the ring of polynomials on countably many indeterminates over $\mathbb{Z}$, because the ideal $I$ generated by the indeterminates isn't finitely generated.
Indeed there is the strictly increasing sequence of ideal $$ (0)\subset (x_1)\subset (x_1,x_2)\subset\dotsb $$ and the union of these ideals is $I$. Why is this sequence strictly increasing? Because if $$ x_{n+1}=x_1f_1+x_2f_2+\dots+x_nf_n \qquad(f_1,f_2,\dots,f_n\in R) $$ evaluating at $x_1=x_2=\dots=x_n=0$ and $x_{n+1}=1$ leads to a contradiction. But if $I$ is finitely generated, its generators belong to one of the ideals in the chain: again a contradiction.
Note that we're talking about finitely generated as modules, not as rings.